# Thread: Find all possible x's for cos(x) >= -1/2

1. ## Find all possible x's for cos(x) >= -1/2

Find all x's for this equation.

$\displaystyle \cos x \geq - \frac{1}{2}$

When I look at the cosine graph, this looks like the answer

$\displaystyle (-\frac{2\pi}{3} + 2\pi k) \leq x \leq (\frac{2\pi}{3} + 2\pi k)$, k is a whole number

but I am not sure. The answer just seems a bit strange. Can someone please tell me if this is the correct way to answer this question?

(Is there a way I can post the cosine graph in this thread?)

Thanks!

2. The unit circle is a good mental tool for this.

cos(x) = -1/2 --> x = 2pi/3

Now, you know that in the interval 0 to 2pi the cosine-function is negative in the second and third quadrants. That is,
cos(2pi/3) = cos(-2pi/3)

The inequality thus holds true for the interval
-2pi/3 < x < 2pi/3
just like you think. Draw a unit circle and plot the values above.

Not sure how used you are to working with trigs, give a shout if you need more clarification.

3. A simple variation on the previous post. $\displaystyle cos(2\pi/3)= -1/2$ and cosine is decreasing so one part of the set is $\displaystyle [0, 2\pi/3$. But cosine is an even function: cos(-x)= cos(x) so $\displaystyle [-2\pi/3, 0]$ will also work- together, $\displaystyle [-2\pi/3, 2\pi/3]$. But, finally, cosine has period $\displaystyle 2\pi$: $\displaystyle cos(x)\ge -1/2$ for all x in $\displaystyle [-2\pi/3+ 2k\pi, 2\pi/3+ 2k\pi]$ for k any integer, exactly what you have!